PN junction built-in potential and depletion width
Enter the doping on the p and n sides and the reverse bias to calculate the built-in potential, the depletion width and how it splits between the two sides, the peak electric field and the junction capacitance, using the depletion approximation for an abrupt junction. You can see in numbers how the depletion region responds to doping and voltage.
Results
A forward voltage beyond the built-in potential removes the depletion region, so this calculation does not apply. Enter a VR greater than −Vbi.
Built-in potential Vbi
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Depletion width W
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Extension into n side xn
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Extension into p side xp
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Peak electric field Emax
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Junction capacitance per area Cj
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Calculated for an abrupt junction with the depletion approximation at 300 K, assuming full ionization. Representative intrinsic carrier densities ni are Si 1.0×10¹⁰, Ge 2.0×10¹³ and GaAs 2.1×10⁶ cm⁻³. Densities can be entered in exponent form such as 1e17. A negative VR means forward bias.
See it as a graph
Other inputs stay at their current values while the chosen input varies from 1/4 to 4 times its current value. The red dot is the current value.
Built-in potential and depletion width
When p-type and n-type regions meet, electrons and holes near the boundary diffuse and recombine, leaving a depletion region of ionized dopants. Their charge sets up the built-in potential Vbi = (kT/q)ln(NA·ND/ni²). Solving Poisson's equation with the depletion approximation, which treats the charge density in the region as constant, gives W = √(2εs(Vbi + VR)/q × (1/NA + 1/ND)). With the defaults of Si, NA = 10¹⁷ cm⁻³, ND = 10¹⁶ cm⁻³ and VR = 0 V, Vbi ≈ 0.774 V and W ≈ 0.332 µm.
The depletion region spreads into the lighter side
The ionized charge on each side must balance, so NA·xp = ND·xn, and the more lightly doped side depletes further. With the defaults the region extends 0.302 µm into the n side but only 0.030 µm into the p side, close to a one-sided junction. The field peaks at the junction, Emax = qND·xn/εs = 2(Vbi + VR)/W, about 46.6 kV/cm here. At a 5 V reverse bias, W grows with the square root of (Vbi + VR) to about 0.906 µm and Emax rises to about 127 kV/cm.
Junction capacitance and C-V measurement
A reverse-biased depletion region acts like a parallel-plate capacitor with charge on both sides, so the capacitance per unit area is Cj = εs/W. It is about 31.2 nF/cm² with the defaults and drops to about 11.4 nF/cm² at a 5 V reverse bias. Varactor diodes use this voltage-dependent capacitance. In a one-sided junction 1/Cj² is linear in voltage, and its slope gives the doping of the lighter side, which is why C-V measurement is used to profile doping.
Materials and design use
Materials with a wider bandgap and smaller ni have a larger built-in potential: at the same doping it is about 0.38 V for Ge and 1.21 V for GaAs. Because ni rises steeply with temperature, Vbi falls as the junction heats up. When reverse bias pushes Emax toward the critical field, roughly 3×10⁵ V/cm for Si depending on doping, avalanche breakdown occurs. Power devices spread the depletion region across a thick, lightly doped drift layer to keep the field low and the breakdown voltage high, while heavy doping on both sides thins W and leads to Zener tunneling breakdown at low voltage.
Under reverse bias the applied voltage adds to the built-in potential, pulling carriers further from the junction, so the depletion region widens. Its width scales roughly as √(Vbi + V), so quadrupling the voltage about doubles it. A wider depletion region lowers the junction capacitance (C = εA/W), the property used in varactor diodes, and spreading the field over a wider region also helps withstand voltage.
Answer: A. It spreads mostly into the lightly doped n side
On both sides of the depletion region the ionized-impurity charge must balance (NA·xp = ND·xn). The lightly doped side must deplete over a wider distance to supply the same charge, so in a p+n junction the depletion region extends almost entirely into the n side. Breakdown voltage and capacitance are then set mostly by the lightly doped side, allowing a one-sided abrupt-junction approximation. Power devices use lightly doped drift layers in this way to hold high voltages.
The built-in potential is Vbi = (kT/q)·ln(NA·ND/ni²), so it grows with doping on both sides. Because the dependence is logarithmic, a tenfold increase in doping adds only about 60 mV, giving typically 0.6–0.9 V in silicon. It cannot be measured with an external voltmeter (contact potentials at the electrodes cancel it), but it guides forward voltage and depletion-width calculations. Wider-gap materials have smaller ni and therefore larger Vbi.
The depletion region behaves like an insulator with charge on both sides, so like a parallel-plate capacitor C = εA/W, and capacitance falls as reverse bias widens W. For example, a 100 µm × 100 µm silicon junction with W = 1 µm has C ≈ 11.7 × 8.85×10⁻¹² × 10⁻⁸ ÷ 10⁻⁶ ≈ 1 pF. Because this capacitance slows high-speed circuits, it is reduced with smaller junction areas or lighter doping.
FAQ
Which way does the depletion region spread in a p+n junction?
Almost entirely into the lightly doped n side. From NA·xp = ND·xn, with NA = 10¹⁹ cm⁻³ and ND = 10¹⁶ cm⁻³, 99.9% of the depletion region lies on the n side, so the breakdown voltage and capacitance are set mainly by the lighter doping.
What happens to the depletion width if the reverse bias is quadrupled?
W is proportional to the square root of (Vbi + VR), so it roughly doubles when VR is much larger than the built-in potential, and the junction capacitance roughly halves. At low voltages where Vbi matters, add it before comparing.
Does the calculator work for forward bias?
Entering a negative VR reduces the potential to Vbi − VF and shows the depletion region shrinking. The depletion approximation becomes inaccurate at high current, and once VF reaches Vbi the width goes to zero, so the formula no longer applies.